A cone (narrow part facing down) is used to contain and transfer a volume of water.
The cone has height $10 m$ and diameter $10 m$.
Initially the cone contains water that takes height up to $8 m$. At one end of the cone there is a small entrance hole so that the water pours down with the rate $10 m^{3}/$min.
Question: How fast does the water elevation in the cone decreases then?
Attempt:
Let $\eta(t)$ be the elevation of the water in the cone at time $t$, and let $V(t)$ be the water volume. By triangle similarities, radius of the water surface in the cone will always be half of the height of water, so let $R(t) = \frac{\eta(t)}{2}$ be the radius of water surface in time. Water volume dynamically changes according to formula $\frac{1}{3} \times base \times height$: $$ V(t) = \frac{1}{3} \pi \frac{\eta^{3}(t)}{4} $$
The draining speed $-10 m^{3}$/min equals $V'(t)$.
$$ -10 = \pi \frac{\eta^{2}(t)}{4} \eta'(t) $$ so implicitly we can say $$ \eta'(t) = -\frac{40}{ \eta^{2} \pi}$$
or solve the following for explicit solution :
$$ \eta^{2} d\eta = -\frac{40}{\pi} dt$$
we get
$$ \frac{1}{3} \eta^{3} = -\frac{40}{\pi}t + C$$ $$ \eta = \left(- \frac{120}{\pi} t + C \right)^{1/3} $$
$$\eta(0) = 8$$, so $C=8^{3}$.
$$ \eta = \left(- \frac{120}{\pi} t + C \right)^{1/3} $$
Then taking derivative we get $$\eta' = \frac{1}{3} \left(-\frac{120}{\pi} \right) \left(- \frac{120}{\pi}t + C\right)^{-2/3} $$
Is there a shorter or better answer compared to this?
I would follow the same steps up to the point where you found this equation:
$$ \eta'(t) = -\frac{40}{\eta^{2}(t) \pi}.$$
As I understand the problem statement, you merely need to find $\eta'$ at the initial conditions, namely $\eta'(0)$. You are given in the problem statement that $\eta(0)=8$. So all you need to do is to plug this value into the equation above.
You will get the same result if you carefully evaluate your final equation at $t=0$, but that takes more effort.